Answer:
b. At twice the distance, the electric potential is V/2
Explanation:
Electric potential
Is the amount of work needed to move a unit charge from a reference point (usually a point at infinity) without producing acceleration.
The electric potential due to a point charge q at a distance r is given by
[tex]\displaystyle V=K\frac{q}{r}[/tex]
Where K is the Coulomb's constant. If we know the electric potential at a certain distance is V, if the distance is changed to 2r, then the new potential is
[tex]\displaystyle V'=K\frac{q}{2r}=\frac{1}{2}K\frac{q}{r}=\frac{1}{2}V[/tex]
It means that the electric potential is half the previous value. Correct option: b.